Integrate (tan x +tan^3 x)/(1+tan^3 x)?

Question

Samantha Adkison · Accepted Answer

#I = 1/6ln|tan^2x - tanx + 1| + 1/sqrt(3)arctan((2tanx - 1)/sqrt(3)) - 1/3ln|tanx+ 1| +C#

The numerator can be rewritten as:

#tanx(1 + tan^2x)#

We know that #tan^2x + 1 = sec^2x#, thus:

#I = int(tanxsec^2x)/(1 + tan^3x)dx#

Let #u = tanx#. Then #du = sec^2xdx# and #(du)/sec^2x = dx#.

#I = int (tanxsec^2x)/(1 + tan^3x) * (du)/sec^2x#

#I = int tanx/(1 + tan^3x) du#

Now we need to change into #u#. This is simple to do because #u = tanx#.

#I =int u/(1 + u^3) du#

We now observe that #u^3 + 1 = (u + 1)(u^2 - u + 1)#.

#I = int u/((u + 1)(u^2 - u + 1)) du#

Finally, using partial fractions, we get:

#A/(u + 1) + (Bu+ C)/(u^2 - u + 1) = u/((u + 1)(u^2 - u + 1))#

#A(u^2 - u + 1) + (Bu + C)(u + 1) = u#

#Au^2 - Au + A + Bu^2 + Cu + B u + C = u#

Write a three-variable system of equations now.

#{(A + B = 0), (B + C - A = 1), (A + C = 0):}#

We can immediately see that #A = -1/3#. This means that #B = C = 1/3#. Thus the integral becomes:

#I = int -1/(3(u + 1)) + (1/3u + 1/3)/(u^2 - u + 1)du#

#I = -1/3ln|u + 1| + 1/3int (u + 1)/(u^2 - u + 1) du#

Now we have another integral to solve. Call it #I_2#.

#I_2 = 1/3int (u + 1)/(u^2 - u + 1)#

We can now rewrite as

#I_2 = 1/3int (1/2(2u - 1) + 3/2)/(u^2 - u + 1) du#

At this point, we can divide into two integrals.

#I_2 = 1/3int(1/2(2u - 1))/(u^2- u + 1) + 3/(2(u^2 - u + 1))#

#I_2 = 1/6 int (2u -1)/(u^2 - u + 1) + 3/(2(u^2 - u + 1)#

Now we let #n = u^2 - u +1#. Then #dn = 2u - 1du# which means that #du= (dn)/(2u - 1)#.

#I_2 = 1/6int (2u - 1)/(u^2 - u + 1) * (dn)/(2u - 1) +int 3/(2(u^2 - u + 1))du#

#I_2 = 1/6int 1/n dn + int 3/(2(u^2 - u + 1))du#

#I_2 = 1/6ln|n| + int 3/(2(u^2 - u + 1))du#

We now have a third integral that needs to be solved. Call it #I_3#.

#I_3 = 3/2int 1/(1(u^2 - u + 1/4 - 1/4) + 1) du#

#I_3 = 3/2int 1/((u - 1/2)^2 + 3/4) du#

#I = 3/2int 1/(((2u - 1)/2)^2 + 3/4)du#

We now let #t = (2u -1)/sqrt(3)#. This signifies that #dt = 2/sqrt(3)du# and that #du = sqrt(3)/2dt#.

#I_3 =(3)/(4sqrt(3))int 1/((sqrt(3)/2t)^2 + 3/4) dt#

#I_3= 3/(4sqrt(3))int 1/(3/4(t^2 + 1)) dt#

#I_3 = 1/sqrt(3)int 1/(t^2 + 1) dt#

#I_3 = 1/sqrt(3)arctan(t)#

Now, the substitutions must be made in reverse.

#I_3 = sqrt(3)arctan((2u - 1)/sqrt(3))#

Reassembling everything, then:

#I = 1/6ln|u^2 - u + 1| + 1/sqrt(3)arctan((2u - 1)/sqrt(3)) - 1/3ln|u + 1| + C#

#I = 1/6ln|tan^2x - tanx + 1| + 1/sqrt(3)arctan((2tanx - 1)/sqrt(3)) - 1/3ln|tanx+ 1| +C#

I hope this is useful!

William Abdalla · Answer

undefined The integral of $\frac{\tan(x) + \tan^3(x)}{1 + \tan^3(x)}$ with respect to $x$ is $\ln|\sec(x)| + C$, where $C$ is the constant of integration.